Answer:
92.04%
Explanation:
Given:
Mass of CO₂ obtained = 53.0 grams
Mass of calcium carbonate heated = 1.31 grams
Now,
the molar mass of the calcium carbonate = 100.08 grams
The number of moles heated in the problem = Mass / Molar mass
= (1.31 grams) / (100.08 grams/moles)
= 0.013088 moles
now,
1 mol of calcium carbonate yields 1 mol of CO₂
thus,
0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂
now,
Theoretical mass of 0.013088 moles of CO₂ will be
= Number of moles × Molar mass of CO₂
= 0.013088 × 44 = 0.5758 grams
Thus, the percent yield for this reaction = 
or
the percent yield for this reaction = 
or
the percent yield for this reaction = 92.04%
Answer:
Thermal energy is a kinetic form of energy that comes from the <u>temperature </u>of matter
Explanation:
Look at the liter man it’s a great way to learn how much every liquid measurement is
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2